Преобразуйте квадрат двучлена в многочлен стандартного вида:
а) $(2\frac{1}{3}a - 1\frac{1}{14}b)^2$;
б) $(0,9x + 1\frac{13}{27}y)^2$;
в) $(-1,2x - 4\frac{1}{6}y)^2$;
г) $(-2,3a + 1\frac{2}{23}b)^2$.
$(2\frac{1}{3}a - 1\frac{1}{14}b)^2 = (\frac{7}{3}a - \frac{15}{14}b)^2 = \frac{49}{9}a^2 - 2 * \frac{7}{3}a * \frac{15}{14}b + \frac{225}{196}b^2 = 5\frac{4}{9}a^2 - \frac{1}{1}a * \frac{5}{1}b + 1\frac{29}{196}b^2 = 5\frac{4}{9}a^2 - 5ab + 1\frac{29}{196}b^2$
$(0,9x + 1\frac{13}{27}y)^2 = (\frac{9}{10}x + \frac{40}{27}y)^2 = \frac{81}{100}x^2 + 2 * \frac{9}{10}x * \frac{40}{27}y + \frac{1600}{729}y^2 = \frac{81}{100}x^2 + 2 * \frac{1}{1}x * \frac{4}{3}y + \frac{1600}{729}y^2 = \frac{81}{100}x^2 + \frac{8}{3}y + \frac{1600}{729}y^2 = \frac{81}{100}x^2 + 2\frac{2}{3}y + 2\frac{142}{729}y^2$
$(-1,2x - 4\frac{1}{6}y)^2 = (-\frac{6}{5}x - \frac{25}{6}y)^2 = \frac{36}{25}x^2 + 2 * \frac{6}{5}x * \frac{25}{6}y + \frac{625}{36}y^2 = 1\frac{11}{25}x^2 + 2 * \frac{1}{1}x * \frac{5}{1}y + 17\frac{13}{36}y^2 = 1\frac{11}{25}x^2 + 10y + 17\frac{13}{36}y^2$
$(-2,3a + 1\frac{2}{23}b)^2 = (-\frac{23}{10}a + \frac{25}{23}b)^2 = \frac{529}{100}a^2 - 2 * \frac{23}{10}a * \frac{25}{23}b + \frac{625}{529}b^2 = 5\frac{29}{100}a^2 - 1 * \frac{1}{1}a * \frac{5}{1}b + 1\frac{96}{529}b^2 = 5\frac{29}{100}a^2 - 5ab + 1\frac{96}{529}b^2$
Пожауйста, оцените решение
